Question

A gas of unknown molar mass was allowed to effuse through a small opening under constant pressure conditions. It required 986 s for 1.00 L of this gas to effuse. Under identical experimental conditions it required 811 s for 1.00 L of chlorotrifluoromethane (CClF3) gas to effuse. Calculate the molar mass of the unknown gas . (Remember that the faster the rate of effusion, the shorter the time required for effusion; that is, rate and time are inversely proportional.)

Answer 1

The molar mass of the unknown gas is 154.4 g/mol

Step-by-step explanation:

Step 1: Data given

It takes 986 seconds for an unknown gas to effuse

It takes 811 seconds for chlorotrifluoromethane (CClF3) gas to effuse

Molar mass of chlorotrifluoromethane (CClF3) = 104.46 g/mol

Step 2: Calculate the molar mass of the gas

t1/t2/ √(M1/M2)

⇒with t1 = the time needed for the unknown gas to effuse = 986 seconds

⇒with t2 = the time needed for CClF3 to effuse = 811 seconds

⇒with M1 = the molar mass of the unknown gas = TO BE DETERMINED

⇒with M2 = the molar mass of CClF3 = 104.46 g/mol

986/811 = √(M1/104.46)

(986/811)² = M1 / 104.46

M1 = 154.4 g/mol

The molar mass of the unknown gas is 154.4 g/mol